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Moni RS Grade: 11
        The general Equation of a conic section is ax^2+2hxy+by^2+2gx+2fy+c=0;


Then how did we get the discriminant as abc+2fgh-af^2-bg^2-ch^2 ??
7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear Moni



The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. Represents a second degree equation where a, h, b doesn’t variables simultaneously.

 

Let a ≠ 0.

 

Now, the above equation becomes

 

        a2 x2 + 2ax (hy + g) = aby2 – 2afy – ac

 

on completing the square on the left side, we get,

 

        a2 x2 + 2ax (hy + g) = y2 (h2 – ab) + 2y (gh – af) + g2 – ac.

 

i.e.    (ax + hy + g) = + √y2(h2–ab)+2y(gh–af) +(g2–ac)

 

We cannot obtain x in terms of y, involving only terms of the first degree, unless the quantity under the radical sign be a perfect square. The condition for this is,

 

(gh – af)2 = (h2 – ab) (g2 – ac)

 

i.e. g2h2 – 2afgh + a2f2 = g2h2 – abg2 – abg2 – ach2 + a2bc

 

        cancelling and diving by a, we have the required condition

 

                abc + 2fgh – af2 – af2 – bg2 – ch2 = 0



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Askiitians Experts
Badiuddin

 

7 years ago
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